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sot
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| geometrycal help required! :( |
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Hello all
Again i need some help with geometry, this time,, i have to make it. The problem is i don't get anything of it :( . I was wondering, if maybe some of u guys can help me? i rly don't anything of it. I made it in paint, as it is in my workbook (scanner doesn't work). Ill put the exercise here:
----Exercise----
Draw all points which just lie as far from A as from B.
Draw all points which lie just as far from the straight CD as the straight EF.
--Picture--
Hooray! if anyone could help me, and if its right, ill give him a reward if he likes to (GW reward)
***THIS POST HAS BEEN EDITED***
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| 05/08/08 10:23 |
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lani
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| RE: geometrycal help required! :( |
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Ehm...
No picture, and shouldn't this be in Off Topic?
Anyhoo, if you draw a line from A to B, then intersect that line at the middle with a line at an 90 degree angle, you'll have all points that are equally distant to both points.
The second part of your question needs graphical representation. I.e. that missing pic.
Ah, there's the pic for the 2nd part.
Draw a line from C to E and one from D to F.
Draw a line through each of the middles of those lines and you have the line of points equally distant from the two earlier lines.
Alternatively, extend the CD and EF lines until they meet. Measure the angle and divide that by two. Draw another line with that angle in relation to EF and you'll end up with the same line.
***THIS POST HAS BEEN EDITED***
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| 05/08/08 10:31 |
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sot
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| RE: geometrycal help required! :( |
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Oh, sorry my bad :( im just back from school excursion (if its called like that). i re uploaded the pic. Soz 2 for my bad english
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| 05/08/08 10:34 |
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Aeon
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| RE: geometrycal help required! :( |
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For part 1, get a compass, open it up so it is slightly over half way between A and B, then put the needle on A and draw an arc above and below the imaginary line between them. Do the same from B. You should end up with 2 crosses. Using a ruler, draw a straight line through these 2 crosses. This gives you the exact answer, with no human error caused by the halfway point not being a whole integer millimetre.
Feel free to pm me ingame if you have any Qs about what I say or about game mechanics, or just wanna talk random crap :p
Main IGNs: Megido Hax / X Megido X
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| 05/08/08 10:35 |
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sot
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| RE: geometrycal help required! :( |
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| QUOTE | Ehm...
No picture, and shouldn't this be in Off Topic?
Anyhoo, if you draw a line from A to B, then intersect that line at the middle with a line at an 90 degree angle, you'll have all points that are equally distant to both points.
The second part of your question needs graphical representation. I.e. that missing pic.
Ah, there's the pic for the 2nd part.
Draw a line from C to E and one from D to F.
Draw a line through each of the middles of those lines and you have the line of points equally distant from the two earlier lines.
Alternatively, extend the CD and EF lines until they meet. Measure the angle and divide that by two. Draw another line with that angle in relation to EF and you'll end up with the same line. |
Sweet! Thx m8, im drawing it right now :D
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| 05/08/08 10:35 |
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lani
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| RE: geometrycal help required! :( |
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| QUOTE | | For part 1, get a compass, open it up so it is slightly over half way between A and B, then put the needle on A and draw an arc above and below the imaginary line between them. Do the same from B. You should end up with 2 crosses. Using a ruler, draw a straight line through these 2 crosses. This gives you the exact answer, with no human error caused by the halfway point not being a whole integer millimetre. |
This will get you two points that are equally far from both points [u]and that distance being the same as the distance between A and B[/b]. This way you're missing out on a whole lot of points equally distant from both A and B where the distance isn't nessecarily the same as the distance between A and B.
Corrction. I didn't read that porperly :-) You're quite right.
***THIS POST HAS BEEN EDITED***
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| 05/08/08 10:39 |
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sot
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| RE: geometrycal help required! :( |
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i don't get this part:
Anyhoo, if you draw a line from A to B, then intersect that line at the middle with a line at an 90 degree angle, you'll have all points that are equally distant to both points.
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| 05/08/08 10:40 |
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Aeon
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| RE: geometrycal help required! :( |
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For Part 2, I think you do this:
Extend the 2 lines until they intersect. At the point of intersection, use a compass to draw a circle. At the points where the circle crosses the 2 lines, draw 2 more circles using those points as the origins (they must be the same size). You should now have 2 points where they intersect, which are a straight line to the point where the 2 lines crossed. Draw a straight line through these 3 points.
Feel free to pm me ingame if you have any Qs about what I say or about game mechanics, or just wanna talk random crap :p
Main IGNs: Megido Hax / X Megido X
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| 05/08/08 10:41 |
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lani
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| RE: geometrycal help required! :( |
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| QUOTE | i don't get this part:
Anyhoo, if you draw a line from A to B, then intersect that line at the middle with a line at an 90 degree angle, you'll have all points that are equally distant to both points. |
Try the other solution. You don't have to understand about angles for that. Besides, do you use 360 degrees angles for a complete circle, or 200 degrees?
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| 05/08/08 10:42 |
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sot
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| RE: geometrycal help required! :( |
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360, but the problem is, i don't understand what u said, im not neglish :( . I Drawed the line from A to B and at the middle of the line, i made a lil dot. What i do then?
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| 05/08/08 10:43 |
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Aeon
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| RE: geometrycal help required! :( |
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| QUOTE | | QUOTE | i don't get this part:
Anyhoo, if you draw a line from A to B, then intersect that line at the middle with a line at an 90 degree angle, you'll have all points that are equally distant to both points. |
Try the other solution. You don't have to understand about angles for that. Besides, do you use 360 degrees angles for a complete circle, or 200 degrees? |
Its 360 degrees, 2π radians, or 400 grades.
Feel free to pm me ingame if you have any Qs about what I say or about game mechanics, or just wanna talk random crap :p
Main IGNs: Megido Hax / X Megido X
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| 05/08/08 10:44 |
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Aeon
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| RE: geometrycal help required! :( |
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| QUOTE | | 360, but the problem is, i don't understand what u said, im not neglish :( . I Drawed the line from A to B and at the middle of the line, i made a lil dot. What i do then? |
EDIT: oops, thought you were referring to his line. From the line between A and B, at the point you made, use a protracter to measure 90 degrees. Then draw a line through the point at 90 degrees, and the original point.
***THIS POST HAS BEEN EDITED***
Feel free to pm me ingame if you have any Qs about what I say or about game mechanics, or just wanna talk random crap :p
Main IGNs: Megido Hax / X Megido X
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| 05/08/08 10:45 |
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sot
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| RE: geometrycal help required! :( |
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:O lol, im stuck with part 1. I did part 2, pretty easy, thx alot u guys :-)
Lets go for part 1 ;D
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| 05/08/08 10:47 |
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Aeon
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| RE: geometrycal help required! :( |
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| QUOTE | :O lol, im stuck with part 1. I did part 2, pretty easy, thx alot u guys :-)
Lets go for part 1 ;D |
The answer is a line that is perpendicular to the line between A and B. In other words, if you tilt the page to an angle where AB is a horizontal line, the 2 lines together will look like a perfect + shape.
Feel free to pm me ingame if you have any Qs about what I say or about game mechanics, or just wanna talk random crap :p
Main IGNs: Megido Hax / X Megido X
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| 05/08/08 10:50 |
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sot
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| RE: geometrycal help required! :( |
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| QUOTE | | QUOTE | :O lol, im stuck with part 1. I did part 2, pretty easy, thx alot u guys :-)
Lets go for part 1 ;D |
The answer is a line that is perpendicular to the line between A and B. In other words, if you tilt the page to an angle where AB is a horizontal line, the 2 lines together will look like a perfect + shape. |
hm, this is what i got atm:
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| 05/08/08 10:57 |
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lani
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| RE: geometrycal help required! :( |
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Almost there.
The line touching the A-B line should intersect it, I.e. go through.
Points along that line but below the A-B line are still equally distant from both points.
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| 05/08/08 10:59 |
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sot
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| RE: geometrycal help required! :( |
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| QUOTE | Almost there.
The line touching the A-B line should intersect it, I.e. go through.
Points along that line but below the A-B line are still equally distant from both points. |
like:
Wich are the points that are equal to both?
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| 05/08/08 11:02 |
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Aeon
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| RE: geometrycal help required! :( |
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| QUOTE | | QUOTE | Almost there.
The line touching the A-B line should intersect it, I.e. go through.
Points along that line but below the A-B line are still equally distant from both points. |
like:
Wich are the points that are equal to both? |
The entire line you are constructing. You should draw it to the edges of the page.
Feel free to pm me ingame if you have any Qs about what I say or about game mechanics, or just wanna talk random crap :p
Main IGNs: Megido Hax / X Megido X
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| 05/08/08 11:04 |
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sot
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| RE: geometrycal help required! :( |
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meh? why? the excersice is 25% of the page, the other rest are exercises.
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| 05/08/08 11:06 |
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sot
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| RE: geometrycal help required! :( |
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but what i draw nwo to get the points that are equal to A and B, that are lieing as far from eachother
---------------------------------------------------
Ooh! the line i drawed in the line of A-B , that line is are the points that are equal, right? i mean, on that line, all the point are equal
***THIS POST HAS BEEN EDITED***
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| 05/08/08 11:07 |
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lani
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| RE: geometrycal help required! :( |
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| QUOTE | | QUOTE |
Wich are the points that are equal to both? |
The entire line you are constructing. You should draw it to the edges of the page. |
Yep. Pick any point along those lines and measure the distance to either A & B for part one, or the original lines in part B. You'll find the distance will be equal. Eyeballing it will work too, but for your understanding of the excersize, should teacher ask you to explain, doing a bit of measuring will do better.
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| 05/08/08 11:08 |
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sot
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| RE: geometrycal help required! :( |
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so, now this exercise is done? :D
it feels GODLIKE!
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| 05/08/08 11:10 |
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sot
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| RE: geometrycal help required! :( |
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meeh >< just found this is my agenda, i need to make it 2
can u help with it 2? this is just text
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| 05/08/08 11:14 |
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sot
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| RE: geometrycal help required! :( |
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bump :P
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| 05/08/08 11:40 |
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sot
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| RE: geometrycal help required! :( |
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bump
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| 05/08/08 13:12 |
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bo fairfieild
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| RE: geometrycal help required! :( |
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| QUOTE | Hello all
Again i need some help with geometry, this time,, i have to make it. The problem is i don't get anything of it :( . I was wondering, if maybe some of u guys can help me? i rly don't anything of it. I made it in paint, as it is in my workbook (scanner doesn't work). Ill put the exercise here:
----Exercise----
Draw all points which just lie as far from A as from B.
Draw all points which lie just as far from the straight CD as the straight EF.
--Picture--
Hooray! if anyone could help me, and if its right, ill give him a reward if he likes to (GW reward) |
For the first part, it is a line whose slope is a negative reciprocal of the slope of a line between AB, and intersects at the midpt of AB.
For the second one, continue line CD and line EF until you have an intersection( let's call it pt G). Draw a line with a slope that =(slope of CD + slope of EF)/2, going through pt G
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| 05/08/08 13:28 |
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sot
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| RE: geometrycal help required! :( |
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| QUOTE | | QUOTE | Hello all
Again i need some help with geometry, this time,, i have to make it. The problem is i don't get anything of it :( . I was wondering, if maybe some of u guys can help me? i rly don't anything of it. I made it in paint, as it is in my workbook (scanner doesn't work). Ill put the exercise here:
----Exercise----
Draw all points which just lie as far from A as from B.
Draw all points which lie just as far from the straight CD as the straight EF.
--Picture--
Hooray! if anyone could help me, and if its right, ill give him a reward if he likes to (GW reward) |
For the first part, it is a line whose slope is a negative reciprocal of the slope of a line between AB, and intersects at the midpt of AB.
For the second one, continue line CD and line EF until you have an intersection( let's call it pt G). Draw a line with a slope that =(slope of CD + slope of EF)/2, going through pt G |
Thx but i was bumping for my second 1, 2 posts above urs ;-)
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| 05/08/08 14:19 |
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dalgur
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| RE: geometrycal help required! :( |
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| QUOTE | meeh >< just found this is my agenda, i need to make it 2
can u help with it 2? this is just text
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A triangle is rectangular if: a²+b²=c²
a and b being the sides next to the rigth angle an c is te side opposite the corner.
So if [AB]²+[AC]²=[BC]² your triangle is rectangular.
T calculate the length of the sides you also need pythagoras.
eg: [AB]²= 1²+2² = 1+4 = 5
[AB] = the square root from 5
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| 05/08/08 14:48 |
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